Electronic – reading the frequency of a sine wave

Any waveform other than a sine at 12.5 MHz will be composed of the fundamental frequency 12.5 MHz, plus harmonics above that. For ideal waveforms with sharp edges, the harmonics extend to infinity.

It sounds like you have added a filter with a cutoff frequency around 12.5 MHz, so this is why your non-sinusoidal outputs don't look like they should -- your filter is stripping off the harmonics which give those waves their non-sinusoidal identity.

To illustrate, here are a few animations from Wikipedia which show various non-sinusoidal waveforms as harmonics are added/removed:

What your filter should be doing is stripping off the harmonics generated by the impulses at each sample output by your DAC. This is called an anti-aliasing filter, and ideally it's a brick-wall low-pass filter with a cutoff equal to half your sampling frequency. Given that true brick-wall filters aren't realizable, the cutoff is usually a bit below half the sampling frequency, and the roll-off is made as steep as feasible. If you can sample significantly more than twice the highest frequency of interest, then the filter design isn't as critical.

It also follows that if you want to synthesize a 12.5 MHz square (or triangle, or sawtooth...) wave, you need a DAC sampling fast enough to also synthesize the higher harmonics in that wave. How much higher depends on how closely you want to approach an ideal square/triangle/sawtooth wave.

The problem with analysing it formulaicly is that the diodes are very non-linear devices and this is implied in the diagram below (red curve is output): -

The blue curve shows a full-wave rectified signal with no capacitor on the output, just a load resistor. When you add a capacitor it charges very quickly due to the forward conduction of one of the diodes but as soon as it attains near-peak voltage (and the blue waveform turns back down towards zero volts), the capacitor remains charged thus reverse biasing the diode and no further diode conduction is seen until the next peak.

You have to think along these lines to adequately "model" what happens. The red curve shows a discharging element between peaks and this is the effect of drawing current into the load resistor. It's a reasonable approximation to regard this as a time-linear discharge but in fact it's an exponential discharge. However, for most decently designed (and used) bridge circuits, using a linear approximation is fine and the ripple voltage formula works pretty good.